1. For a particle, the de-Broglie wavelength is

$\lambda = \frac{h}{p}$

where $h$ is Planck’s constant and $p$ is momentum.

2. Since all particles have the same kinetic energy $K$, use the non-relativistic relation:

$K = \frac{p^2}{2m}$

So,

$p = \sqrt{2mK}$

Substituting into de-Broglie relation:

$\lambda = \frac{h}{\sqrt{2mK}}$

3. Because $h$ and $K$ are same for all particles,

$\lambda \propto \frac{1}{\sqrt{m}}$

Thus, smaller mass means larger wavelength.

4. Compare masses:

• Electron mass $m_e$ is smallest
• Proton mass $m_p$ is much larger than electron
• $\alpha$-particle mass $m_\alpha \approx 4m_p$, so it is the largest

Therefore,

$m_e < m_p < m_\alpha$

Hence,

$\lambda_e > \lambda_p > \lambda_\alpha$

5. Checking options:

• A: incorrect
• B: incorrect
• C: correct
• D: incorrect

Therefore, the correct option is:

$\boxed{\lambda_e > \lambda_p > \lambda_\alpha}$